# Expand a number

Your task is to expand an integer greater than 0 in the following manner:

Split the number into decimal digits and for each digit generate a list according to these rules:

• if the digit is odd, the list starts with the digit and goes down to 1;
• if the digit is even, the list starts with the digit and goes up to 9.
• if the digit is 0, the list is empty.

Write down the lists for the odd numbers below the digits, and above for the even ones. Then top-align the columns and collect the digits in each row to make integers. As a final step add up the numbers to find the expansion of the number.

Here's an example of the above rules applied to 34607:

 9
8
79
68
57
346 7 ->  399 7 -> 3997 -> 9418
2   6     288 6    2886
1   5     177 5    1775
4      66 4     664
3      5  3      53
2      4  2      42
1         1       1


Here are the test cases:

1: 1
2: 44
3: 6
44: 429
217: 1270
911: 947
2345: 26114
20067: 3450
34875632: 70664504
9348765347634763: 18406119382875401


This is , so the shortest answers in bytes in each language win.

• Can we take input as a string? Or as an array of digits? Nov 8 '17 at 21:56
• @Arnauld It must be an integer, and to split it to digits with your program/function Nov 8 '17 at 21:59
• @GalenIvanov But if input is from stdin, I suppose that's ok (even though technically a string), right?
Nov 8 '17 at 22:05
• @Adám Yes, technically it's a string, so that's ok. Nov 9 '17 at 6:03
• And here I thought it would be an expansion like this. Nov 9 '17 at 14:29

# Jelly, 13 bytes

Dḟ0RrḂ?€9UZḌS


Try it online! or see the test-suite.

### How?

Dḟ0RrḂ?€9UZḌS - Link: positive integer           e.g. 702
D             - cast to a decimal list                [7,0,2]
0           - literal zero                          0
9     - literal nine
€      - for each:
?       -   if:
Ḃ        -   ...condition: bit (modulo by 2)      1              ,0
R          -   ...then: range ([1,...n])            [1,2,3,4,5,6,7],n/a
r         -   ...else: inclusive range ([n,...9])  n/a            ,[2,3,4,5,6,7,8,9]
U    - upend                                 [[7,6,5,4,3,2,1],[9,8,7,6,5,4,3,2]]
Z   - transpose                             [[7,9],[6,8],[5,7],[4,6],[3,5],[2,4],[1,3],2]
Ḍ  - cast from decimal lists               [79,68,57,46,35,24,13,2]
S - sum                                   324


# Perl 6,  68  66 bytes

{sum roundrobin(.comb».&{$_%2??($_...1)!!(9...+$_) if +$_})».join}


Try it

{sum roundrobin(.comb».&{[R,] $_%2??1..$_!!$_..9 if +$_})».join}


Try it

## Expanded:

{  # bare block lambda with implicit parameter ｢$_｣ sum roundrobin( # grab from the sub lists in a round robin fashion .comb\ # split the input into digits ».&{ # for each digit do this [R,] # reduce with reversed &infix:«,» (shorter than reverse)$_ % 2     # is the digit not divisible by 2?
??  1  .. $_ # from 1 to the digit !!$_ ..  9   # from the digit to 9

if +$_ # only do the above if the current digit isn't 0 # (has the effect of removing 0 from the list) } )».join # join each of the sub-lists from roundrobin }  # APL (Dyalog), 39 bytes Full program body. Prompts for input from STDIN. Prints result to STDOUT. +/10⊥¨0~⍨¨↓⍉↑{2|⍵:⌽⍳⍵⋄×⍵:⌽⍵,⍵↓⍳9⋄⍬}¨⍎¨⍞  Try it online! To display the last test case correctly, ⎕FR (Floating-point Representation) has been set to 128-bit Decimal and ⎕PP (Print Precision) has been set to 34 digits. ⍞ prompt for text input from STDIN ⍎¨ execute each (gets each digit as a number) {}¨ for each element, apply the following function where the argument is represented by ⍵: 2|⍵: if odd (lit. "if" division remainder when divided by 2), then: ⌽ reverse ⍳ the ɩntegers from 1 until ⍵ the argument ⋄ else ×⍵: if the argument is positive (lit. "if" signum), then: ⌽ reverse ⍵ the argument , followed by ⍵ argument ↓ elements dropped from ⍳9 the the ɩntegers from 1 until 9 ⋄ else ⍬ empty list ↑ mix (combine) this list of lists into a single matrix, padding with zeros on the right ⍉ transpose ↓ split this matrix into a list of lists 0~⍨¨ remove all zeros from each list 10⊥¨ convert each from base-10 to normal numbers (this collects the digits) +/ sum the numbers • Thank you for your explanation. Just to let you know that I discovered code golf after watching you in the APL code golf video. Nov 9 '17 at 6:13 # JavaScript (ES6), 8883 82 bytes f=(n,k=0)=>k<9&&+[...n+''].map(x=>+x&&(x=x&1?x:9-k<x||9)>k?x-k:'').join+f(n,k+1)  ### Note 9 - k < x || 9 saves a byte over 9 - k >= x && 9 but generates 1 instead of 0 if the inequality is verified. It would be a problem if it leaded to 1 > k, triggering the wrong path in the outer ternary. But it would mean that k = 0 and therefore 9 - k = 9, so we can't possibly have 9 - k < x at the same time. ### Test cases NB: Removed the last test case which exceeds JS number precision. f=(n,k=0)=>k<9&&+[...n+''].map(x=>+x&&(x=x&1?x:9-k<x||9)>k?x-k:'').join+f(n,k+1) console.log(f(1)) // 1 console.log(f(2)) // 44 console.log(f(3)) // 6 console.log(f(44)) // 429 console.log(f(217)) // 1270 console.log(f(911)) // 947 console.log(f(2345)) // 26114 console.log(f(20067)) // 3450 console.log(f(34875632)) // 70664504 # Japt, 18 bytes ì f ®òZu ª9 wÃy xì  Try it online! # Java 11, 210209191 181 bytes n->{long r=0;var a="0".repeat(9).split("");for(int d:(n+"").getBytes())for(int k=0,x=d-48,y=9;x>0&(k<1||(d%2<1?y-->x:x-->1));a[k++]+=d%2<1?y:x);for(var q:a)r+=new Long(q);return r;}  Ok, this took quite a while (mainly because I made a mistake at first, so had to write down each step to better understand what I did wrong). Can most likely be golfed some more. -18 bytes thanks to @ceilingcat. Explanation: Try it here. n->{ // Method with long as both parameter and return-type long r=0; // Result-long r, starting at 0 var a="0".repeat(9).split(""); // String array a, filled with nine String zeroes for(int d:(n+"").getBytes()) // Cast the input to a String, // and loop over its codepoints as integers: for(int k=0, // Row-index k, starting at x=d-48, // Temp integer x, set to the current digit y=9 // Temp integer y, set to 9 ; // Inner loop, if: x>0 // The current digit is not a 0, &(k<1 // and if this is the first iteration, ||(d%2<1? // or if the digit is even: y-->x // And y is larger than the digit // (and afterwards decrease y by 1 with y--) : // or if the digit is odd: x-->1)); // And x is larger than 1 // (and afterwards decrease x by 1 with x--) a[k++]+= // Append the current row with: // (and afterwards increase k by 1 with k++) d%2<1? // If the digit is even: y // Append the row with y : // Else (the digit is odd): x); // Append the row with x for(var q:a) // Loop over the String rows in the array: r+=new Long(q); // Convert it to a long, and add it to the result-sum return r;} // Return the result  # Pip, 28 bytes J_MS(RV{a?a%2?\,aa,tl}Ma)ZDx  Try it online! ### Explanation  Ma Map this function to the digits of the 1st cmdline arg: a? If digit != 0: a%2? If digit is odd: \,a Inclusive-range(digit) a,t Else (digit is even), range from digit to 10 (exclusive) l Else (digit is 0), empty list RV{ } Apply reverse to the result before returning it This gives us a list of lists like [9 8 7 6] or [3 2 1] ( )ZDx Zip, with a default value of empty string J_MS Use map-sum to join each sublist and sum the results Autoprint (implicit)  How the steps proceed with 34607 as the argument: 34607 [[1 2 3] [4 5 6 7 8 9] [6 7 8 9] [] [1 2 3 4 5 6 7]] [[3 2 1] [9 8 7 6 5 4] [9 8 7 6] [] [7 6 5 4 3 2 1]] [[3 9 9 "" 7] [2 8 8 "" 6] [1 7 7 "" 5] ["" 6 6 "" 4] ["" 5 "" "" 3] ["" 4 "" "" 2] ["" "" "" "" 1]] [3997 2886 1775 664 53 42 1] 9418  # Pyth - 23 bytes siRT.Tm*!!d@,}9d}d1dsMz  # Haskell, 106 104 bytes import Data.List f n=sum$map read$transpose$[reverse$[[c..'9'],['1'..c]]!!mod(read[c])2|c<-show n,c>'0']  Try it online! # R, 153 146 bytes function(n,m=n%/%10^(nchar(n):0)%%10)sum(strtoi(apply(sapply(m[m>0],function(x)c(r<-"if"(x%%2,x:1,9:x),rep("",9-sum(r|1)))),1,paste,collapse="")))  Try it online! Sometimes, I can't tell if I'm just garbage at golfing, or if R is.... It's definitely me, saved 7 bytes thanks to user2390246, who reminded me of another way to extract digits (that I suggested myself). You may replace strtoi with as.double to get 18406718084351604 for the last test case (which is wrong); R only has 32-bit integers. • You can save a bit by taking n as an integer and converting to digits using one of your tricks! 146 bytes Nov 9 '17 at 13:50 • @user2390246 you know, I tried that, but I think I was fixated on using the implicit conversion to character->int when I called : and then used strtoi anyway! Nov 9 '17 at 14:10 # Perl 5, 120 + 1 (-a) = 121 bytes $p=y/01357/ /r;$n=y/02468/ /r;map{$p=~s/9/ /g;$p=~s/\d/$&+1/ge;$n=~s/\d/$&-1/ge;$n=~s/0/ /g;@F=($p,@F,\$n)}0..7;say for@F


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# Python 2, 131 bytes

lambda n:sum(int(''.join(nfor n in l if n))for l in map(None,*[range(n and(n%2*n or 9),(n%2<1)*~-n,-1)for n in map(int,n*10)]))


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# 05AB1E, 16 bytes

0KεDÈi9ŸëL]íõζJO


Or alternatively:

0KεDÈ8*>Ÿ{R}õζJO
0Kε9Ÿ¬L‚yèR}õζJO


Explanation:

0K        # Remove all 0s from the (implicit) input-integer
ε       #  Map each digit to:
D      #   Duplicate the digit
Èi    #   If it's even:
9Ÿ  #    Pop and push a list in the range [digit, 9]
ë    #   Else (the digit is odd):
L   #    Pop and push a list in the range [1, digit]
]       # Close both the if-else statement and map
í      # Reverse each inner ranged list
ζ    # Zip/transpose, swapping rows and columns,
õ     # with an empty string as filler
J   # Join each inner list together
O  # And sum that list
# (after which the result is output implicitly)


# Burlesque, 39 bytes

riXX:nz{J2dv{{9r@}{ro}}che!<-}m[tp)im++


Try it online!

ri      #Read as int
XX      #Return list of digits
:nz     #Filter for non-zero
{
J2dv   #Divisible by 2?
{
{9r@} #Range a, 9
{ro}  #Range 1, a
}che!  #Run based on if divisible
<-     #Reverse the range
}m[     #Apply to each digit
tp      #Transpose digits
)im     #Join each list into single int
++      #Sum each int